3 added 14 characters in body

Depending on what you consider simple, let $k$ be the complex numbers, or the integers, or the field with two elements (or any other commutative ring you're fond of). Let $R=k[a,b,c,x,y,z]/(ax+by+cz-1)$. Map $R^3$ to $R$ by $(f,g,h)\mapsto xf+yg+zh$. Let $P$ be the kernel of this map.

$P$ is the universal example of a rank 2 projective module over a $k$-algebra that becomes free after adding a free rank-one direct summand. (That is, if $A$ is any other $k$-algebra with such a projective module $Q$, then there is a map from $R$ to $A$ such that $Q=P\otimes_RA$.) One could argue that this makes it the simplest example. In particular, Hugh Thomas's example arises from this example in this way.

To see that $P$ is not free, use the fact that Hugh Thomas's example is not free. Alternatively, invoke the much more general result of Mohan Kumar and Nori, which says that if $R=k[x_1,...,x_n,a_1,...a_n]/(\sum x_ia_i-1)$, then the kernel of the map defined by the $1\times n$ matrix $(x_1^{m_1},\ldots x_n^{m_n})$ cannot be free unless $m_1\ldots m_n$ is divisible by $(n-1)!$.

If you want an example where the UFD property is obvious, Mohan Kumar's paper "On Stably Free Modules" gives a family of examples over rings of the form $A_f$ where the rings $A$ are polynomial rings over fields. These examples all have the property that they become free after adding a free rank-one module.

Depending on what you consider simple, let $k$ be the complex numbers, or the integers, or the field with two elements (or any other commutative ring you're fond of). Let $R=k[a,b,c,x,y,z]/(ax+by+cz-1)$. Map $R^3$ to $R$ by $(f,g,h)\mapsto xf+yg+zh$. Let $P$ be the kernel of this map.

$P$ is the universal example of a rank 2 projective module over a $k$-algebra that becomes free after adding a direct summand. (That is, if $A$ is any other $k$-algebra with such a projective module $Q$, then there is a map from $R$ to $A$ such that $Q=P\otimes_RA$.) One could argue that this makes it the simplest example. In particular, Hugh Thomas's example arises from this example in this way.

To see that $P$ is not free, use the fact that Hugh Thomas's example is not free. Alternatively, invoke the much more general result of Mohan Kumar and Nori, which says that if $R=k[x_1,...,x_n,a_1,...a_n]/(\sum x_ia_i-1)$, then the kernel of the map defined by the $1\times n$ matrix $(x_1^{m_1},\ldots x_n^{m_n})$ cannot be free unless $m_1\ldots m_n$ is divisible by $(n-1)!$.

If you want an example where the UFD property is obvious, Mohan Kumar's paper "On Stably Free Modules" gives a family of examples over rings of the form $A_f$ where the rings $A$ are polynomial rings over fields. These examples all have the property that they become free after adding a free rank-one module.

1

Depending on what you consider simple, let $k$ be the complex numbers, or the integers, or the field with two elements (or any other commutative ring you're fond of). Let $R=k[a,b,c,x,y,z]/(ax+by+cz-1)$. Map $R^3$ to $R$ by $(f,g,h)\mapsto xf+yg+zh$. Let $P$ be the kernel of this map.

$P$ is the universal example of a rank 2 projective module over a $k$-algebra that becomes free after adding a direct summand. (That is, if $A$ is any other $k$-algebra with such a projective module $Q$, then there is a map from $R$ to $A$ such that $Q=P\otimes_RA$.) One could argue that this makes it the simplest example. In particular, Hugh Thomas's example arises from this example in this way.

To see that $P$ is not free, use the fact that Hugh Thomas's example is not free. Alternatively, invoke the much more general result of Mohan Kumar and Nori, which says that if $R=k[x_1,...,x_n,a_1,...a_n]/(\sum x_ia_i-1)$, then the kernel of the map defined by the $1\times n$ matrix $(x_1^{m_1},\ldots x_n^{m_n})$ cannot be free unless $m_1\ldots m_n$ is divisible by $(n-1)!$.